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7 tháng 6 2017

Ta có :

A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

A = \(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

A = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

A = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

A =  \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Tách A thành 2 nhóm,ta được :

A = \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Lại có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\text{ }\text{ }\)

            \(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\text{ }\text{ }\)

A > \(\left(\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{75}.25+\frac{1}{100}.25\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

A < \(\left(\frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}\right)+\left(\frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}\right)=\frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25\)

\(=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

7 tháng 6 2017

Bạn SKT_NTT làm đúng rồi nha

7 tháng 6 2017

mình vừa mới trả lời xong đấy 

Câu hỏi của Do Not Ask Why - Toán lớp 7 - Học toán với OnlineMath

NV
24 tháng 6 2019

\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{60}\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{50}+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

2/ \(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=\frac{7}{12}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}>\frac{7}{12}\)

Tương tự câu trên ta có: \(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(A=\frac{1}{51}+...+\frac{1}{60}+\frac{1}{61}+...+\frac{1}{70}+\frac{1}{71}+...+\frac{1}{80}+\frac{1}{81}+...+\frac{1}{90}+\frac{1}{91}+...+\frac{1}{100}\)

\(A< \frac{1}{50}+...+\frac{1}{50}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{70}+...+\frac{1}{70}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{90}+...+\frac{1}{90}\)

\(A< 10.\frac{1}{50}+10.\frac{1}{60}+10.\frac{1}{70}+10.\frac{1}{80}+10.\frac{1}{90}\)

\(A< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{5}{6}\)

29 tháng 12 2019

+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{9900}\)

\(A=\left(\frac{1}{2}+\frac{1}{12}\right)+\left(\frac{1}{30}+...+\frac{1}{9900}\right)>\frac{1}{2}+\frac{1}{12}.\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{12}\)

\(\Rightarrow A>\frac{7}{12}\left(1\right).\)

+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{5}{6}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}< \frac{5}{6}\)

\(\Rightarrow A< \frac{5}{6}\left(2\right).\)

Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\left(đpcm\right).\)

Chúc bạn học tốt!

Ta thấy : A = 1 - 1/2 + 1/3 - 1/4 +...+ 1/99 +1/100

             A = 1 + 1/2 + 1/3 + 1/4 +...+ 1/99 + 1/100 - 2. (1/2 + 1/4 +1/6 +...+ 1/100)

            A = 1 + 1/2 + 1/3 +1/4 +...+ 1/99 + 1/100 - (1 + 1/2 + 1/3 +...+1/50)

            A = 1/51 + 1/52 + 1/53 +...+ 1/100

Do đó : A = (1/51 + 1/52 + 1/53 +...+ 1/75) + (1/76 + 1/77 +...+ 1/100)

Ta có : 1/51 > 1/52 > ... > 1/75    ;     1/76 > 1/77 > ... > 1/100 nên :

           A > 1/75.25 + 1/100.25 = 1/3 + 1/4 = 7/12

           A < 1/51.25 < 1/50.25 + 1/75.25 = 1/2 +1/3 = 5/6

Vậy 7/12 < A <5/6 . ^_^

9 tháng 10 2021

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\text{ CMR : }\frac{7}{12}< A< \frac{5}{6}\)

Ta có :

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2.\frac{1}{2}-2.\frac{1}{4}-...-2.\frac{1}{98}\)

\(A=1+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{49}\)

\(A=\frac{1}{51}+...+\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{51.25}=\frac{25}{51}< \frac{25}{30}=\frac{5}{6}\) (đpcm)

Và \(A>25.\frac{1}{75}+25.\frac{1}{100}=\frac{7}{12}\)

Ta có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)

Lại có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100)

                =  (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <  1 - 1 / 2 + 1 / 3 = 5 / 6 (2)

Từ (1) và (2) => 7 / 12 < A < 5 / 6

7 tháng 9 2015

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)

=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)